2001 AIME I Problems/Problem 7
Triangle has , and . Points and are located on and , respectively, such that is parallel to and contains the center of the inscribed circle of triangle . Then , where and are relatively prime positive integers. Find .
Let be the incenter of , so that and are angle bisectors of and respectively. Then, so is isosceles, and similarly is isosceles. It follows that , so the perimeter of is . Hence, the ratio of the perimeters of and is , which is the scale factor between the two similar triangles, and thus . Thus, .
The semiperimeter of is . By Heron's formula, the area of the whole triangle is . Using the formula , we find that the inradius is . Since , the ratio of the heights of triangles and is equal to the ratio between sides and . From , we find . Thus, we have
Solving for gives so the answer is .
Or we have the area of the triangle as . Using the ratio of heights to ratio of bases of and from that it is easy to deduce that .
Solution 3 (mass points)
Using the angle bisector theorem, we know the ratio is , thus we shall assign a weight of to point and a weight of to point , giving a weight of . In the same manner, using another bisector, we find that has a weight of . So, now we know has a weight of , and the ratio of is . Therefore, the smaller similar triangle is the height of the original triangle . So, is the size of . Multiplying this ratio by the length of , we find is . Therefore, .
Solution 4 (Faster)
More directly than Solution 2, we have
Diagram borrowed from Solution 3.
Let the angle bisector of intersects at .
Applying the Angle Bisector Theorem on we have Since is the angle bisector of , we can once again apply the Angle Bisector Theorem on which gives Since we have Solving gets . Thus .
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